Braids, orderings and minimal volume cusped hyperbolic 3-manifolds
It is well-known that there is a faithful representation of braid groups on automorphism groups of free groups, and it is also well-known that free groups are bi-orderable. We investigate which -strand braids give rise to automorphisms which preserve some bi-ordering of the free group of rank . As a consequence of our work we find that of the two minimal volume hyperbolic -cusped orientable -manifolds, one has bi-orderable fundamental group whereas the other does not. We prove a similar result for the -cusped case, and have further results for more cusps. In addition, we study pseudo-Anosov braids and find that typically those with minimal dilatation are not order-preserving.
Key words and phrases:ordered groups, bi-orderings, knots and links, braid groups, free groups, mapping class groups, fibered 3-manifolds
2010 Mathematics Subject Classification:Primary 20F60, 57M07, Secondary 57M50, 57M27,
If is a strict total ordering of the elements of a group such that implies for all , we call a left-ordered group. If the left-ordering is also invariant under right-multiplication, we call a bi-ordered group (sometimes known in the literature simply as “ordered” group). If a group admits such an ordering it is said to be left- or bi-orderable. It is easy to see that a group is left-orderable if and only if it is right-orderable. If is a bi-ordered group, then is invariant under conjugation: if and only if for all . Nontrivial examples of bi-orderable groups are the free groups of rank , as discussed in Appendix A.
An automorphism of a group is said to preserve an ordering of if for every we have ; we also say is -invariant. If preserves an ordering of , then the th power preserves the ordering of for each integer . We note that if preserves an ordering it does not necessarily follow that does.
E. Artin [2, 3] observed that each -strand braid corresponds to an automorphism of . This paper concerns the question of which braids give rise to automorphisms which preserve some bi-ordering of . In turn this is related to the orderability of the fundamental group of the complement of certain links in , namely the braid closure together with its axis, which we call a braided link. We pay special attention to pseudo-Anosov mapping classes and their stretch factors (dilatations), and cusped hyperbolic -manifolds of small volume.
This paper is organized as follows: Section 2 reviews the relation among braids, mapping class groups, free group automorphisms and certain links in the 3-sphere . In Section 3 we recall basic properties of orderable groups, with explicit bi-ordering of free groups further described in Appendix A. The study of order-preserving braids and their relation to bi-ordering the group of the corresponding braided links is initiated in Section 4. Applications to cusped hyperbolic 3-manifolds of minimal volume are considered in Section 5. In Section 6 we give many examples of non-order-preserving braids, including pseudo-Anosov braids with minimal dilation as well as large dilatations. We also find a family of pretzel links whose fundamental groups can not be bi-orderable. Appendix B is devoted to a proof that the fundamental group of the Whitehead link complement is bi-orderable.
Acknowledgments. We thank Tetsuya Ito for helpful conversations and comments.
2. Braids and
Let be the -strand braid group, which has the well-known presentation with generators subject to the relations if and if . See Figure 1(1)(2).
2.1. Mapping classes
Let denote the disk with punctures, which we may picture as equally spaced along a diameter of the disk and labelled to . denotes the mapping class group of and the mapping class group of homeomorphisms fixed on the boundary pointwise. There is a well-known isomorphism
which sends to a half twist which interchanges the punctures labelled and , see Figure 1(3). The kernel of the obvious map is infinite cyclic, generated by a Dehn twist along a simple closed curve parallel to the boundary of the disk. Using the isomorphism together with this obvious map, we have the surjective homomorphism
whose kernel is generated by the full twist , where is the half twist.
Elements of are classified into three types: periodic, reducible and pseudo-Anosov, called Nielsen-Thurston types . If two given mapping classes are conjugate to each other, then their Nielsen-Thurston types are the same. We say that is periodic (resp. reducible, pseudo-Anosov) if its mapping class is of the corresponding type.
2.2. Free group automorphisms
Let be an -strand braid. Let be a representative of the mapping class . Obviously represents a mapping class . If one passes to the induced map of the fundamental group of , using a point on the boundary as basepoint, this defines the Artin representation
which can be defined on the generators as follows, where are the free generators of the free group of rank and is the group of automorphisms of . The generator induces the automorphism
See Figure 2. It is known that the Artin representation is faithful. Its image is the subgroup of automorphisms of that take each to a conjugation of some and which take the product to itself.
We note that if for -strand braids and , then for some integer . The images of and under the Artin representation are the same up to an inner automorphism
By abuse of notation, from now on we will use the same symbol for the braid, the mapping class and the corresponding automorphism of .
An automorphism of is said to be symmetric if for each generator , the image is a conjugate of some . Every automorphism on corresponding to the action of a braid on is symmetric. A symmetric automorphism is pure if sends each to a conjugate of itself. Pure braids (see Section 4.2) induce symmetric and pure automorphisms.
Since braid words, like paths, are typically read from left to right, we adopt the convention that braids act on on the right. If , we denote the action of by , and if we have the identity .
An -strand braid is said to be order-preserving if there exists some bi-ordering of preserved by the automorphism of .
One sees that is order-preserving if and only if is order-preserving for some (hence all) (Corollary 4.2).
As discussed below, there is some ambiguity in defining the action of on , depending on choices of a representative of the mapping class , basepoint in and generators of . As we’ll see, this ambiguity is irrelevant in the question of whether a braid is order-preserving.
It is sometimes convenient to use a basepoint and generators different from that used in the Artin representation. Specifically, we consider a representative of a mapping class in and we may assume that are two different points (not necessarily on the boundary of the disk), each fixed by . Then we have induced maps and . Of course, and are isomorphic, but not canonically. We can construct an isomorphism by choosing a path in from to which then defines an isomorphism sending the class of a loop in based at to the class of the loop based at . Consider the diagram
which is not necessarily commutative. However, we leave it to the reader to check the following
The above diagram commutes up to conjugation. Specifically, equals followed by a conjugation in , the conjugating element being the class of the loop in .
The map preserves a bi-ordering of if and only if preserves a bi-ordering of
If preserves the bi-ordering of , define a bi-ordering of by the formula for . Then one checks that using conjugation invariance of bi-orderings. The converse is proved similarly. ∎
If one passes from to , there is a further ambiguity regarding the action of a braid on . However, this ambiguity corresponds to conjugation by a power of , so again it is irrelevant to the question of preserving a bi-ordering. More concretely, given an -strand braid , let be any representative of the mapping class . We take any basepoint of possibly . By choosing a path in from to , we have the induced map
which sends the class of a loop in based at to the class of the loop based at the same point. By using Corollary 2.3, one sees that is order-preserving if and only if preserves a bi-ordering of .
By abuse of notation again, we denote the induced map by , when is specified.
These remarks show that if one allows different choices of basepoint, say a basepoint , the action of on should really be regarded as a representation
the group of outer automorphisms. Recall that , where is the (normal) subgroup of inner automorphisms of a group .
2.4. Mapping tori and braided links
For a braid , we denote the mapping torus by
where is a representative of . By the hyperbolization theorem of Thurston , is hyperbolic if and only if is pseudo-Anosov.
The closure of a braid is a knot or link in the -sphere and the braided link, denote by , is the closure , together with the braid axis , which is an unknotted curve that runs around in a monotone manner: . See Figure 3(1)(2). Whereas all links can be realized as , this is not true for , as each component of has nonzero linking number with the braid axis . As an example the Whitehead link considered in Appendix B is not a braided link.
We see that is homeomorphic with the complement of the braid closure in the solid torus . The interior can be identified with the complement of in , so we have the following.
is homeomorphic to .
Of course the fundamental group of is isomorphic with that of , which in turn is the semidirect product
There is a (split) exact sequence
One may consider as the set of ordered pairs
with the multiplication given by
3. Orderable groups
A group is left-orderable if and only if there exists a sub-semigroup such that for every exactly one of , or holds.
Indeed such a defines a left-ordering by the rule . Conversely, a left-ordering defines a positive cone satisfying the conditions of Proposition 3.1.
A group is bi-orderable if and only if it posesses as in Proposition 3.1, and in addition for all .
We note that a left- or bi-ordering of is preserved by an automorphism if and only if , where is the positive cone of the ordering.
Suppose is an exact sequence of groups. If and are left-ordered with positive cones and respectively, then is left-orderable using the positive cone .
This is sometimes called the lexicographic order of , considered as an extension.
In Proposition 3.3, if and are bi-ordered, then the formula defines a bi-ordering of if and only if the bi-ordering of is respected under conjugation by elements of ; equivalently for all .
A subset of a left-ordered group is said to be convex if for all and satisfying , we have .
If is an exact sequence of groups with a left-ordered group and a convex subgroup of , then there is a left-ordering of in which if and only if some (and hence every) element satisfies . If is a bi-ordered group, so is .
Being bi-orderable is a much stronger property than being left-orderable. An intermediate property of groups is to be locally-indicable, which means that every finitely-generated nontrivial subgroup has the infinite cyclic group as a quotient.
For a group, the following implications hold: bi-orderable locally-indicable left-orderable torsion-free. None of these implications is reversible.
Proposition 3.7 ().
If is a knot or link in , then its group is locally indicable and therefore left-orderable.
Certain knot groups and link groups are bi-orderable, and this is a particular focus of this paper. For example, torus knot groups are not bi-orderable, but the figure-eight knot has bi-orderable group , and we shall see that the same is true of the Whitehead link and many links constructed from braids as above. One of the reasons bi-orderability of knot groups is of interest has to do with surgery and -spaces, which were introduced by Ozsváth and Szabó  and include all 3-manifolds with finite fundamental group.
Theorem 3.8 ().
If is a knot in for which is bi-orderable, then surgery on cannot produce an -space.
We note that this is not true in general for links. As we shall see in Theorem B.1, the Whitehead link has bi-orderable fundamental group, but one can construct lens spaces by certain surgeries on the link.
Consider a group , an automorphism and the semidirect product . Recall with multiplication given by
Suppose is bi-orderable and is an automorphism. Then is bi-orderable if and only if there exists a bi-ordering of which is preserved by .
Suppose that is bi-ordered by . Since bi-orderings are invariant under conjugation, the equation implies that , restricted to , is -invariant. For the converse, suppose preserves a bi-ordering of . Then we use the lexicographic ordering defined by if and only if as integers or and , to order . It is easily checked that this is a bi-ordering using the identity and the assumption that preserves the ordering . ∎
Let be an orientable surface. Then is bi-orderable, see . Let be the mapping class group of , let and assume for some . Since the fundamental group of the mapping torus of is the semidirect product , where is an automorphism induced from , we have the following from Proposition 3.9.
is bi-orderable if and only if there exists a bi-ordering of which is preserved by , where is a representative of the mapping class .
If for , then we say that the orbit of (under ) is trivial.
Let be a left-orderable group. If an automorphism preserves a left-ordering of , then cannot have any nontrivial finite orbits.
If the orbit of is nontrivial, we may assume . Then , and by transitivity and induction for all and therefore the orbit of , , is infinite. ∎
Proposition 3.11 says that if has a nontrivial finite orbit, then does not preserve any left-ordering of .
Consider an automorphism of a free group. The following two criteria for being order-preserving (or not) will be useful; they involve the abelianization and its eigenvalues, which are, a priori, complex numbers, possibly with multiplicity.
Theorem 3.12 ().
Let be an automorphism. If every eigenvalue of is real and positive, then there is a bi-ordering of which is -invariant.
Theorem 3.13 ().
If there exists a bi-ordering of which is -invariant, then has at least one real and positive eigenvalue.
This is useful in showing that certain fibred -manifolds have fundamental groups which are not bi-orderable. However, we note that in the case of braids it does not apply in that way, for if is an automorphism of induced by a braid , or more generally a symmetric automorphism, then is simply a permutation of the generators of and therefore has at least one eigenvalue equal to one.
We note that Theorem 3.12 cannot have a full converse. It has been observed in  that for there exist automorphisms of which preserve a bi-ordering of , but whose eigenvalues are precisely the th roots of unity in . Such examples appear in the discussion in Section 4.6.
In Appendix A a certain class of bi-orderings of the free group , called standard orderings is defined, using the lower central series.
If is a non-pure symmetric automorphism, then cannot preserve any standard bi-ordering of .
Assume is not pure, but preserves a standard bi-ordering of . Then is a nontrivial permutation of the generators of the abelianization . By Proposition A.1, the commutator subgroup is convex relative to , and therefore inherits a bi-ordering according to Proposition 3.5. Since preserves , one easily checks that preserves the order of . But, being non-pure symmetric implies that is a nontrivial permutation of the generators of which are the images of the . By Proposition 3.11, cannot preserve any ordering of , a contradiction. ∎
If is a pure symmetric automorphism, then is order-preserving. In fact, it preserves every standard ordering of .
A pure symmetric automorphism induces the identity map , so Proposition A.3 applies. ∎
For any symmetric automorphism , there exists so that is pure symmetric. By Proposition 3.15, we have the following.
If is a symmetric automorphism, then there exists such that is order-preserving.
4. Order-preserving braids
4.1. Basic properties
A braid is order-preserving if and only if is bi-orderable.
Let be the -strand braid
and let be the half twist
The full twist is written by , which means that and are roots of . We note that commutes with all -strand braids and in fact generates the centre of when , see [24, Theorem 1.24].
A braid is order-preserving if and only if is order-preserving for some (hence all) Moreover, they preserve exactly the same bi-orderings of .
This follows since and are homeomorphic: By using the disk twist as we shall define in Section 4.4, we see that th power of the disk twist about the disk bounded by the braid axis of sends the exterior of the link to the exterior . An alternative argument is that acts by conjugation of , so it preserves every bi-order of . ∎
As noted by Garside  every -strand braid has an expression where is a Garside positive braid word, meaning that can be written as a word in the generators without negative exponents. Thus a question of a braid being order-preserving can always be reduced to the case of positive braids. Notice that changing a braid by conjugation does not change the link up to isotopy, so we have:
Let . Then is order-preserving if and only if is order-preserving.
The very simplest of nontrivial braids are the generators .
The generators are not order-preserving.
Suppose that preserves a bi-order of . We may assume . Then , or in other words , see (2.1). But conjugation invariance of the ordering then yields , which is a contradiction.
An alternative argument is as follows. We take a basepoint in the interior of . There exists a homeomorphism which represents such that the induced map gives rise to the following automorphism on :
see Figure 4(2). (In the figures, we denote by , the image of under the automorphism of , and denote by , the inverse of .) Suppose that preserves a bi-order of . We may assume . Then since preserves . This implies that by conjugation invariance of the bi-ordering. This is a contradiction. ∎
4.2. Pure braids
As is well-known, there is a homomorphism of the -strand braid group onto the permutation group of letters, and the kernel is the group of pure braids .
Every pure braid is order-preserving, and in fact preserves every standard bi-order of .
For a pure braid , the image of the Artin representatation is pure symmetric. This completes the proof by Proposition 3.15. ∎
The nd power of each generator is a pure braid. Thus we have the following.
The braids are order-preserving.
The 2-strand braid gives rise to the examples of links in Figure 6 whose complements have bi-orderable fundamental groups. All the link complements are homeomorphic to one another. It is clear that they are homeomorphic with , whose fundamental group is isomorphic with .
By Corollary 3.16, we immediately obtain the following.
For every braid some power is order-preserving. The fundamental group is bi-orderable and may be regarded as a normal subgroup of index in .
4.3. Periodic braids
Suppose that is a periodic braid. It is known (see for example [17, page 30]) that there exists an integer such that is conjugate to either
Let be a periodic braid.
If is conjugate to for some , then is order-preserving and is bi-orderable.
If is conjugate to for some with , then is not order-preserving and can not be bi-orderable.
Note that is order-preserving when since . Theorem 4.10 is a consequence of [7, Theorem 1.5]: If is a periodic braid, then is a Seifert fibered -manifold. If is of type 1, then has no exceptional fibres. If is of type and if is not a multiple of , then has an exceptional fibre.
4.4. Disk twists
We review a method to construct links in whose complements are homeomorphic to each other. Let be a link in . We denote a tubular neighborhood of by , and the exterior of , that is by . Suppose that contains an unknot as a sublink. Then (resp. ) is homeomorphic to a solid torus (resp. torus). We denote the link by . Taking a disk bounded by the longitude of , we define two homeomorphisms
as follows. We cut along . We have resulting two sides obtained from . Then we reglue the two sides by rotating either of the sides degrees so that the mapping class of the restriction defines the right-handed Dehn twist about , see Figure 7(1). Such an operation defines the homeomorphism . If segments of pass through , then is obtained from by adding a full twist braid near . In the case , see Figure 7(2). For example, if is equivalent to for some and if is taken to be the braid axis of , then is equivalent to the closure of . Notice that determines the latter homeomorphism
We call the (right-handed) disk twist about .
For any integer , we have a homeomorphism of the th power so that is the th power of the right-handed Dehn twist about . Observe that converts into a link in such that is homeomorphic to . We denote by , a homeomorphism: and call the th power of (right-handed) disk twist about .
4.5. Alternative proof of Theorem 4.10
Proof of Theorem 4.10.
We prove the claim (1). Consider the pure -strand braid . Then is a link consisting of three unknotted components, including the axis. Performing a disk twist times on one of the components of the closed braid converts the braided link of the 2-strand braid to the braided link of the -strand braid , which is conjugate to the type 1 braid , see Figure 8. But the disk twist being a homeomorphism of the complement of the link, we see that . But since is pure, has bi-orderable fundamental group, in fact isomorphic with Hence the fundamental group of is bi-orderable and is order-preserving. Thus the th power is also order-preserving.
We turn to the claim (2). We use the basepoint in the interior of . There exists a homeomorphism which represents such that the induced map gives rise to the following automorphism111The product of ’s in (4.1) is given by , if . This is equal to in (4.2) up to an inner automorphism. on :
The half twist is order-preserving if and only if is odd.
If , then is conjugate to , and if , then is conjugate to with . By Theorem 4.10, we finish the proof. ∎
As a special case of Theorem 4.10(2), we see that the periodic 3-strand braid is not order-preserving. Another way to see this is to observe that the 3-strand braid gives the following automorphism on by using the Artin representation
One can show this doesn’t preserve a bi-order of as follows. Assume some bi-order were preserved by the map, then supposing without loss of generality that , we would have (applying ), hence (conjugation invariance), so (applying ) and then (transitivity) and finally the contradiction (conjugation again).
On the other hand is pseudo-Anosov, and in fact it is the simplest pseudo-Anosov -strand braid; see Section 6.1. We thank George Bergman for pointing out the following argument. It will also follow from more general result in Section 6.2; see Corollary 6.4.
The braid is not order-preserving.
Let be the free generators of . Using the Artin representation , one sees that the action of is
Consider the orbit of the element under this action. Assuming there is a bi-order of invariant under this action, we may assume without loss of generality that , and therefore all elements of the orbit of are positive. Moreover implies and since we have .
Now the calculation
and the facts that and are negative show that the action is decreasing on the orbit of , which is also the orbit of .
Calculating the preimages of the generators, we have the action of expressed as